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# If x and y are positive integers such that x=8y+12, what is

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If x and y are positive integers such that x=8y+12, what is [#permalink]  05 Jun 2005, 17:43
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If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y?
1) x=12u, where u is an integer.
2) y=12z, where z is an integer.
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Re: DS: GCD [#permalink]  05 Jun 2005, 18:03
Vithal wrote:
If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y?
1) x=12u, where u is an integer.
2) y=12z, where z is an integer.

1) y = (x - 12)/8 = 12(u-1)/8 and 12u so u-1 must be divisible by 2, 12 may not be a GCD, the GDC for these two numbers is at least 2, but depending on u it can be greater than 2.
say u-1 is divisible by 8 then GDC can be 12.
so insufficient

2) y=12z and x = 12(8z+1), here GCD is 12, is there a GCD >1 for z and 8z+1? no because
suppose there is a divisor m such that z/m = integer then 8z/m + 1/m is not.

so GDC is 12 and sufficient.

B.
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Re: DS: GCD [#permalink]  05 Jun 2005, 21:21
sparky wrote:
Vithal wrote:
If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y?
1) x=12u, where u is an integer.
2) y=12z, where z is an integer.

1) y = (x - 12)/8 = 12(u-1)/8 and 12u so u-1 must be divisible by 2, 12 may not be a GCD, the GDC for these two numbers is at least 2, but depending on u it can be greater than 2.
say u-1 is divisible by 8 then GDC can be 12.
so insufficient

2) y=12z and x = 12(8z+1), here GCD is 12, is there a GCD >1 for z and 8z+1? no because
suppose there is a divisor m such that z/m = integer then 8z/m + 1/m is not.

so GDC is 12 and sufficient. B.

sparky, agree with you. but more than this you have nice approach to handle the problem. thanx. keep up the sprit.
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Re: DS: GCD [#permalink]  30 Oct 2011, 12:19
sparky wrote:
Vithal wrote:
If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y?
1) x=12u, where u is an integer.
2) y=12z, where z is an integer.

1) y = (x - 12)/8 = 12(u-1)/8 and 12u so u-1 must be divisible by 2, 12 may not be a GCD, the GDC for these two numbers is at least 2, but depending on u it can be greater than 2.
say u-1 is divisible by 8 then GDC can be 12.
so insufficient

2) y=12z and x = 12(8z+1), here GCD is 12, is there a GCD >1 for z and 8z+1? no because
suppose there is a divisor m such that z/m = integer then 8z/m + 1/m is not.

so GDC is 12 and sufficient.

B.

i do not understand why 12 cant be gcd of statement 1 ... Can someone please explain ??? and I dont understand the reasoning behind
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Re: DS: GCD [#permalink]  31 Oct 2011, 04:59
x=8y+12
statement 1
x=12u

12u=8y+12; y=3/2(u-1). Given y is a positive integer, so u=3,5,7...; y=3,6,9...
For y=3; x=36. GCD(3,36) is 3.
For y=6; x=60. GCD(6,60) is 6. Insufficient

Statement 2
y=12z
x=8*12z+12 or 12(8z+1). Both x and y are multiples of 12. Therefore GCD is 12. Sufficient
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Re: DS: GCD [#permalink]  31 Oct 2011, 08:41
How can we say GCD is 12? Couldn't it be 24?
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Re: DS: GCD [#permalink]  31 Oct 2011, 10:09
Mindreko wrote:
How can we say GCD is 12? Couldn't it be 24?

That is not possible
for any value of z, y and x both can not be multiple of 24

Posted from my mobile device
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Re: DS: GCD   [#permalink] 31 Oct 2011, 10:09
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